Ravi Kant Kumar

If \[ T_n=\sin^n x+\cos^n x \] Prove that \[ 2T_6-3T_4+1=0 \] Solution: \[ T_6=\sin^6 x+\cos^6 x \] \[ = (\sin^2 x+\cos^2 x)^3 -3\sin^2 x\cos^2 x(\sin^2 x+\cos^2 x) \] \[ = 1-3\sin^2 x\cos^2 x \] Also, \[ T_4=\sin^4 x+\cos^4 x \] \[ = (\sin^2 x+\cos^2 x)^2 -2\sin^2 x\cos^2 x \] \[ = 1-2\sin^2 x\cos^2 x \]

If \[ T_n=\sin^n x+\cos^n x \] Prove that \[ \frac{T_3-T_5}{T_1} = \frac{T_5-T_7}{T_3} \] Solution: \[ T_3-T_5 = (\sin^3 x+\cos^3 x) -(\sin^5 x+\cos^5 x) \] \[ = \sin^3 x(1-\sin^2 x) + \cos^3 x(1-\cos^2 x) \] \[ = \sin^3 x\cos^2 x + \cos^3 x\sin^2 x \] \[ = \sin^2 x\cos^2 x(\sin x+\cos x) \] \[ = \sin^2

Prove that \[ \sqrt{\frac{1-\sin x}{1+\sin x}} + \sqrt{\frac{1+\sin x}{1-\sin x}} = -\frac{2}{\cos x} \] where \[ \frac{\pi}{2}

If \[ a=\sec x-\tan x \] and \[ b=\cosec x+\cot x \] Show that \[ ab+a-b+1=0 \] Solution: \[ a=\sec x-\tan x = \frac{1-\sin x}{\cos x} \] \[ b=\cosec x+\cot x = \frac{1+\cos x}{\sin x} \] Now, \[ ab+a-b+1 \] \[ = \frac{(1-\sin x)(1+\cos x)} {\sin x\cos x} + \frac{1-\sin x}{\cos x} – \frac{1+\cos x}{\sin

If \[ \sin x+\cos x=m \] Prove that \[ \sin^6 x+\cos^6 x = \frac{4-3(m^2-1)^2}{4} \] where \[ m^2\le 2 \] Solution: \[ \sin^6 x+\cos^6 x \] \[ = (\sin^2 x)^3+(\cos^2 x)^3 \] \[ = (\sin^2 x+\cos^2 x) (\sin^4 x-\sin^2 x\cos^2 x+\cos^4 x) \] \[ = \sin^4 x+\cos^4 x-\sin^2 x\cos^2 x \] \[ = (\sin^2 x+\cos^2

If \[ \cot x(1+\sin x)=4m \] and \[ \cot x(1-\sin x)=4n \] Prove that \[ (m^2-n^2)^2=mn \] Solution: \[ 4m=\cot x(1+\sin x) = \frac{\cos x}{\sin x}(1+\sin x) \] \[ 4n=\cot x(1-\sin x) = \frac{\cos x}{\sin x}(1-\sin x) \] Multiplying, \[ 16mn = \frac{\cos^2 x}{\sin^2 x}(1-\sin^2 x) \] \[ = \frac{\cos^4 x}{\sin^2 x} \] Now, \[

If \[ \cosec x-\sin x=a^3,\qquad \sec x-\cos x=b^3 \] Prove that \[ a^2b^2(a^2+b^2)=1 \] Solution: \[ \cosec x-\sin x = \frac{1}{\sin x}-\sin x = \frac{1-\sin^2 x}{\sin x} = \frac{\cos^2 x}{\sin x} \] \[ a^3=\frac{\cos^2 x}{\sin x} \] Similarly, \[ \sec x-\cos x = \frac{1}{\cos x}-\cos x = \frac{1-\cos^2 x}{\cos x} = \frac{\sin^2 x}{\cos x} \]

If \[ \tan x=\frac{a}{b} \] Show that \[ \frac{a\sin x-b\cos x} {a\sin x+b\cos x} = \frac{a^2-b^2}{a^2+b^2} \] Solution: Since \[ \tan x=\frac{a}{b} \] \[ \sin x=\frac{a}{\sqrt{a^2+b^2}} \] and \[ \cos x=\frac{b}{\sqrt{a^2+b^2}} \] Now, \[ \frac{a\sin x-b\cos x} {a\sin x+b\cos x} \] \[ = \frac{ a\left(\frac{a}{\sqrt{a^2+b^2}}\right) – b\left(\frac{b}{\sqrt{a^2+b^2}}\right) } { a\left(\frac{a}{\sqrt{a^2+b^2}}\right) + b\left(\frac{b}{\sqrt{a^2+b^2}}\right) } \] \[

If \[ \tan x=\frac{b}{a} \] Find the Value of \[ \sqrt{\frac{a+b}{a-b}} + \sqrt{\frac{a-b}{a+b}} \] Solution: \[ \sqrt{\frac{a+b}{a-b}} + \sqrt{\frac{a-b}{a+b}} \] Taking LCM, \[ = \frac{ (a+b)+(a-b) } { \sqrt{(a+b)(a-b)} } \] \[ = \frac{2a}{\sqrt{a^2-b^2}} \] Since \[ \tan x=\frac{b}{a} \] \[ 1-\tan^2 x = 1-\frac{b^2}{a^2} = \frac{a^2-b^2}{a^2} \] \[ \sqrt{a^2-b^2} = a\sqrt{1-\tan^2 x} \] Therefore,

If \[ \sin x=\frac{a^2-b^2}{a^2+b^2} \] Find the Values of \[ \tan x,\ \sec x \text{ and } \cosec x \] Solution: \[ \sin x=\frac{a^2-b^2}{a^2+b^2} \] Using \[ \sin^2 x+\cos^2 x=1 \] \[ \cos x = \sqrt{1-\sin^2 x} \] \[ = \sqrt{ 1- \left( \frac{a^2-b^2}{a^2+b^2} \right)^2 } \] \[ = \sqrt{ \frac{(a^2+b^2)^2-(a^2-b^2)^2} {(a^2+b^2)^2} } \] \[